Solve for x: 2 over 5 (x − 2) = 4x
2 answers:
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I’m pretty sure this is right:
Suppose that some value, c, is a point of a local minimum point.
The theorem states that if a function f is differentiable at a point c of local extremum, then f'(c) = 0.
This implies that the function f is continuous over the given interval. So there must be some value h such that f(c + h) - f(c) >= 0, where h is some infinitesimally small quantity.
As h approaches 0 from the negative side, then:
As h approaches 0 from the positive side, then:
Thus, f'(c) = 0
The theorem states that if a function f is differentiable at a point c of local extremum, then f'(c) = 0.
This implies that the function f is continuous over the given interval. So there must be some value h such that f(c + h) - f(c) >= 0, where h is some infinitesimally small quantity.
As h approaches 0 from the negative side, then:
As h approaches 0 from the positive side, then:
Thus, f'(c) = 0